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October 7, 2020, 23:58 
multiphase interfoam komega BC

#1 
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Diego Villegas Díaz
Join Date: Oct 2020
Posts: 4
Rep Power: 4 
Hi, I am a new foamer. I am simulating a section of an openchanel steady water flow with interFoam solver.
The inlet condition is a water profile and a constant flow in. To represent this situation I used this boundary condition: 1) alpha.water: I used funkySetBoundaryFields to map the waterprofile to alpha.water variable. 1 for water and 0 for air, without intermediate values. 2) U: I used the variableHeightFlowRateInletVelocity with the inlet flow. 3) p_rgh: zero gradient. 4) nut: calulated. 5) k: turbulentIntensityKineticEnergyInlet; with Intensity = 0.0573 6) omega: turbulentMixingLengthFrequencyInlet; with mixing length = 0.00258 I've had some troubles with the k and omega boundary conditions. interFoam fall at the start and I dont understand why. Meanwhile, I am using a fixed value for omega (3.46) or fixed value for k (0.0044065), and its work but I really want to understand why when I use the boundary condition 5) and 6) together interFoam fall. Thaks in advance. 

October 8, 2020, 02:08 

#2 
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Charles
Join Date: Aug 2016
Location: Vancouver, Canada
Posts: 138
Rep Power: 8 
I'm curious how do you get these values, i.e. intensity 0.0573, mixing length 000258, omega 3.46 and k 0.0044065. I guess the divergence was caused by improper initial conditions because turbulence model is prone to diverge.


October 8, 2020, 09:53 

#3  
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Diego Villegas Díaz
Join Date: Oct 2020
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Quote:
"Turbulence in open channel flows" (Nezu & Nakagawa) the mixing length scale, k = (3/2)*(I*U)^2, the mixing length for the omega condition I calculated as 0.07*Rh (with Rh the hydraulic radius) I found this scale in these forum but I dont have any formal information. Finally to calculate omega (for fixed value), I calculated another length scale to mixing length lm equal to 0.41*Rh (based in Von Karman constant) and omega is equal to sqrt(k)/lm. I forgot to mention that my system is initialized at the inlet as uniforms values for k, omega and lm, the same values that I mentioned above. May be these unifor initial condition is the reason why interFoam falls. 

October 8, 2020, 21:10 

#4 
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Charles
Join Date: Aug 2016
Location: Vancouver, Canada
Posts: 138
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k = (3/2)*(I*U)^2 is not mixing length, but is turbulent kinetic energy. So, I guess you need to review these initial values.


October 8, 2020, 23:41 

#5 
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Diego Villegas Díaz
Join Date: Oct 2020
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Ohh I made a mistake writing the last post. I know that k is turbulent Kinect energy. But I'm thinking that interFoam fall due the uniform initial condition un k and omega.


October 9, 2020, 02:44 

#6 
Senior Member
Charles
Join Date: Aug 2016
Location: Vancouver, Canada
Posts: 138
Rep Power: 8 
You have two mixing length. The first mixing length 0.07Rh might be problematic and is too small. You can try 0.41Rh instead.


Tags 
boundary condition, interfoam, komega model, open channel flow 
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